Explanation:
Potential energy stored \(=U\)
\(U = \frac{1}{2}k{x^2}\)
\(U = \frac{1}{2}k \times {(2)^2}\)
\(k = \frac{{2U}}{4} = \frac{U}{2}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{(}}1)\)
After stretching by \(10 \mathrm{~cm} . U^{\prime}=\) new potential energy to \(U^{\prime}=\dfrac{1}{2} k x^{\prime 2}\)
using (1) \(x^{\prime}=\) new distance \(=10 \mathrm{~cm}\)
\( \Rightarrow \frac{1}{2} \times \frac{U}{2} \times {(10)^2}\;\;\;{\mkern 1mu} {\kern 1pt} \left( {k = \frac{U}{2}} \right)\)
\(U' = 25\,U\)